How Do You Calculate Static Friction on an Inclined Plane?

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To calculate static friction on an inclined plane, the scenario involves a 10 kg block on a 45° incline with a downward force of 10 N applied. The normal force (N) is determined by the equation N = mg cos(45°) + 10 N, leading to a value of 79.3 N. The static friction force (Ff) is calculated as 69.3 N, which is less than the maximum static friction force, indicating the block remains at rest. The coefficient of static friction (μs) is derived from the relationship μs = Ff / N, but it is noted that μs is greater than or equal to this ratio. The discussion emphasizes the importance of understanding the conditions under which static friction operates.
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Homework Statement



Suppose a block of mass = 10kg is on an incline plane that makes an angle = 45° from the horizontal. If a force of 10 N is being applied to the block, downward, and perpendicular to the surface, and the block is not moving, what is the static force of friction? What is the coefficient of static friction?



Homework Equations



\mu_{}s = Fn

The Attempt at a Solution




\mu_{}s = FN

Am I on the right track?
 
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laladude said:

Homework Statement



Suppose a block of mass = 10kg is on an incline plane that makes an angle = 45° from the horizontal. If a force of 10 N is being applied to the block, downward, and perpendicular to the surface, and the block is not moving, what is the static force of friction? What is the coefficient of static friction?

3.jpg


Homework Equations



\mu_{}s = Fn

The Attempt at a Solution



ƩFy = ma
ƩFy = N + -mg + - 10N = 0
N = (10N)mg * cos45°
should be N -mgcos45 -10 = 0, N = mgcos45 +10. Don't confuse the normal force, N, with the force unit of Newtons (N)!
friction = mg * sinθ
f = 69.3 N
This is the correct answer for the first part.
\mu_{}s = FN

Am I on the right track?
you mean F_f is less than or equal to \mu_sN, don't you?
 
PhanthomJay said:
should be N -mgcos45 -10 = 0, N = mgcos45 +10. Don't confuse the normal force, N, with the force unit of Newtons (N)!This is the correct answer for the first part.you mean F_f is less than or equal to \mu_sN, don't you?

LOL Yes, I did mean that. So it will be \mu_{}s = Ff / N?

Which will be 69.3N/79.3N..correct?
 
laladude said:
LOL Yes, I did mean that. So it will be \mu_{}s = Ff / N?

Which will be 69.3N/79.3N..correct?
No-o. Since static friction force is always less than or equal to \mu N , then \mu_{}s is greater than or equal to[/color] Ff / N.
 
PhanthomJay said:
No-o. Since static friction force is always less than or equal to \mu N , then \mu_{}s is greater than or equal to[/color] Ff / N.

Oh, okay! Thank you so much :biggrin:
 
You're welcome. Your calculation for the Normal force is also correct. Note that only if the block was at rest but just on the verge of moving would the static friction coefficient be equal to Ff/N. Otherwise, it would be greater than Ff/N.
 
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